LOVASZ-SCHRIJVER SDP-OPERATOR, NEAR-PERFECT GRAPHS …ltuncel/publications/1411.2069.pdf ·...

LOVASZ-SCHRIJVER SDP-OPERATOR,

NEAR-PERFECT GRAPHS AND NEAR-BIPARTITE GRAPHS

S. BIANCHI, M. ESCALANTE, G. NASINI, L. TUNCEL

Abstract. We study the Lovasz-Schrijver lift-and-project operator (LS+) based on the cone of

symmetric, positive semidefinite matrices, applied to the fractional stable set polytope of graphs.

The problem of obtaining a combinatorial characterization of graphs for which the LS+-operator

generates the stable set polytope in one step has been open since 1990. We call these graphs

LS+-perfect. In the current contribution, we pursue a full combinatorial characterization of LS+-

perfect graphs and make progress towards such a characterization by establishing a new, close

relationship among LS+-perfect graphs, near-bipartite graphs and a newly introduced concept

of full-support-perfect graphs.

1. Introduction

The notion of a perfect graph was introduced by Berge in the early 1960s [5]. A graph is called

perfect if each of its induced subgraphs has chromatic number equal to the size of a maximum

cardinality clique in the subgraph. Perfect graphs have caught the attention of many researchers

in the area and inspired numerous very interesting contributions to the literature for the past

fifty years. One of the main results in the seminal paper of Grotschel, Lovasz and Schrijver [20]

is that perfect graphs constitute a graph class where the Maximum Weight Stable Set Problem

(MWSSP) can be solved in polynomial-time. Some years later, the same authors proved a very

beautiful, related result which connects a purely graph theoretic notion to polyhedrality of a

typically nonlinear convex relaxation and to the integrality and equality of two fundamental

polytopes:

Date: November 19, 2014, revised: March 21, 2016.

Key words and phrases. stable set problem, lift-and-project methods, semidefinite programming, integer

programming.

Some of the results in this paper were first announced in conference proceedings in abstracts [7, 8]. This

research was supported in part by grants PID-CONICET 241, PICT-ANPCyT 0361, ONR N00014-12-10049, and

a Discovery Grant from NSERC.

S. Bianchi: Universidad Nacional de Rosario, Argentina (e-mail: [email protected])

M. Escalante: Universidad Nacional de Rosario, Argentina (e-mail: [email protected])

G. Nasini: Universidad Nacional de Rosario, Argentina (e-mail: [email protected])

L. Tuncel: Corresponding author. Department of Combinatorics and Optimization, Faculty of Mathematics,

University of Waterloo, Waterloo, Ontario N2L 3G1, Canada (e-mail: [email protected]).

1

2 S. BIANCHI, M. ESCALANTE, G. NASINI, L. TUNCEL

Theorem 1. (Grotschel, Lovasz and Schrijver [20, 21]) For every graph G, the following are

equivalent:

(1) G is perfect;

(2) STAB(G) = CLIQUE(G);

(3) TH(G) = STAB(G);

(4) TH(G) = CLIQUE(G);

(5) TH(G) is polyhedral.

In the above theorem, STAB(G) is the stable set polytope of G, CLIQUE(G) is its clique

relaxation and TH(G) is the theta body of G defined by Lovasz [27].

In the early 1990s, Lovasz and Schrijver [28] introduced the semidefinite relaxation LS+(G)

(LS+ operator was originally denoted by N+ in [28]) of STAB(G) which is stronger than TH(G).

Following the same line of reasoning used for perfect graphs, they proved that MWSSP can be

solved in polynomial-time for the class of graphs for which LS+(G) = STAB(G). We call these

graphs LS+-perfect graphs (originally, the authors called these graphs N+-perfect [7]). The

set of LS+-perfect graphs is known to contain many rich and interesting classes of graphs (e.g.,

perfect graphs, t-perfect graphs, wheels, anti-holes, near-bipartite graphs) and their clique sums.

However, no combinatorial characterization of LS+-perfect graphs have been obtained so far.

There are many studies of various variants of lift-and-project operators applied to the relax-

ations of the stable set problem (see for instance, [34, 24, 15, 23, 9, 26, 16, 31, 22, 18, 19]). Why

study LS+-perfect graphs? For example, if we want to characterize the largest family of graphs

for which MWSSP can be solved in polynomial-time, then perhaps, we should pick a tractable

relaxation of STAB(G) which is as strong as possible. This reasoning would suggest that, we

should focus on the strongest, tractable lift-and-project operator and reiterate it as much as

possible while maintaining tractability of the underlying relaxation. Even though the (lower

bound) analysis for the strongest lift-and-project operators is typically very challenging, some

considerable amount of work on the behaviour of the strongest lift-and-project operators applied

to the stable set problem already exists (see [1] and the references therein). In the spectrum

of strong lift-and-project operators which utilize positive semidefiniteness constraints, given the

above results of Grotschel, Lovasz and Schrijver [20, 21], it seems clear to us that we should

pick an operator which is at least as strong as TH(G). Among many of the convex relaxations

that are closely related to TH(G) but stronger, TH(G) continues to emerge as the central object

with quite special mathematical properties (see [10] and the references therein). Given that the

operator LS+(G) can be defined as the intersection of the matrix-space liftings of the odd-cycle

polytope of G and the theta body of G, by definition, LS+(G) encodes and retains very interest-

ing combinatorial information about the graph G. Then, the next question is why not focus on

iterated (hence stronger) operator LSk+ for k ≥ 2 but k small enough to maintain tractability?

The answer to this is related to the above; but, it is a bit more subtle: in the lifted, matrix-space

representation of LS+, if we remove the positive semidefiniteness constraint, we end up with the

lifting of the operator LS (defined later). In this lifted matrix space, if we remove the restriction

that the matrix be symmetric, we end up with the lifting of a weaker relaxation LS0. LSk0 retains

many interesting combinatorial properties of G, see [24, 25]. Moreover, Lovasz and Schrijver

proved that for every graph G, LS0(G) = LS(G). However, this property does not generalize to

SDP-OPERATOR, NEAR-PERFECT AND NEAR-BIPARTITE GRAPHS 3

the iterated operators LSk0, LSk, even for k = 2, even if we require that the underlying graph G

be perfect (see [2, 3]). Therefore, LS+ has many of the desired attributes for this purpose.

One of our main goals in this line of research is to obtain a characterization of LS+-perfect

graphs similar to the one given in Theorem 1 for perfect graphs. More precisely, we would like

to find an appropriate polyhedral relaxation of STAB(G) playing the role of CLIQUE(G) in

Theorem 1, when we replace TH(G) by LS+(G). In [7] we introduced the polyhedral relaxation

NB(G) of STAB(G), which is, to the best of our knowledge, the tightest polyhedral relaxation

of LS+(G). Roughly speaking, NB(G) is defined by the family of facets of stable set polytopes

of a family of graphs that are built from near-bipartite graphs by using simple operations so

that the stable set polytope of the resulting graph does not have any facets outside the class of

facets which define the stable set polytope of near-bipartite graphs (for a precise definition of

NB(G), see Section 2). In our quest to obtain the desired characterization of LS+-perfect graphs,

NB(G) is our current best guess for replacing CLIQUE(G) in Theorem 1. More specifically, we

conjecture that the next four statements are equivalent.

Conjecture 2. For every graph G, the following four statements are equivalent:

(1) STAB(G) = NB(G);

(2) LS+(G) = STAB(G);

(3) LS+(G) = NB(G);

(4) LS+(G) is polyhedral.

Verifying the validity of Conjecture 2 is equivalent to determining the validity of the following

two statements:

Conjecture 3. For every graph G, if LS+(G) is polyhedral then STAB(G) = NB(G).

Conjecture 4. For every graph G, if LS+(G) = STAB(G) then STAB(G) = NB(G).

In [6], we made some progress towards proving Conjecture 3, by presenting an infinite family

of graphs for which it holds. Recently, Conjecture 4 was verified for web graphs [17]. In this

contribution, we prove that Conjecture 4 holds for a class of graphs called fs-perfect graphs

that stand for full-support-perfect graphs. This graph family was originally defined in [29] and

includes the set of near-perfect graphs previously defined by Shepherd in [32].

One of the main difficulties in obtaining a good combinatorial characterization for LS+-perfect

graphs is that the lift-and-project operator LS+ (and many related operators) can behave sporad-

ically under many well-studied graph-minor operations (see [16, 26]). Therefore, in the study of

LS+-perfect graphs we are faced with the problem of constructing suitable graph operations and

then deriving certain monotonicity or loose invariance properties under such graph operations.

In this context, we present two operations which preserve LS+-imperfection in graphs.

In the next section, we begin with notation and preliminary results that will be used through-

out the paper. We also state our main characterization conjecture on LS+-perfect graphs. In

Section 3, we characterize fs-perfection in the family of graphs built from a minimally imperfect

graph and one additional node. In Section 4 we prove the validity of the conjecture on fs-perfect

graphs. In order to ease the reading of this contribution, the proofs of results on the LS+


operator are presented in Section 5. Section 6 is devoted to the conclusions and some further

results.

2. Further definitions and preliminary results

2.1. Graphs and the stable set polytope. Throughout this work, G stands for a simple

graph with node set V (G) and edge set E(G). The complementary graph of G, denoted by G, is

such that V (G) := V (G) and, E(G) := {uv : u 6= v, u, v ∈ V (G), uv /∈ E(G)}. For any positive

integer n, Kn, Cn and Pn denote the graphs with n nodes corresponding to a complete graph, a

cycle and a path, respectively. We assume that in the cycle Cn node i is adjacent to node i+ 1

for i ∈ {1, . . . , n− 1} and n is adjacent to node 1.

Given V ′ ⊆ V (G), we say that G′ is a subgraph of G induced by the nodes in V ′ if V (G′) = V ′

and E(G′) = {uv : uv ∈ E(G), {u, v} ⊆ V (G′)}. When V (G′) is clear from the context, we say

that G′ is a node induced subgraph of G and write G′ ⊆ G. Given U ⊆ V (G), we denote by

G−U the subgraph of G induced by the nodes in V (G)\U . For simplicity, we write G−u instead

of G− {u}. We say that GE is an edge subgraph of G if V (GE) = V (G) and E(GE) ⊆ E(G).

Given the graph G, the set ΓG(v) is the neighbourhood of node v ∈ V (G) and δG(v) = |ΓG(v)|.The set ΓG[v] = ΓG(v) ∪ {v} is the closed neighbourhood of node v. When the graph is clear

from the context, we simply write Γ(v) or Γ[v]. If G′ ⊆ G and v ∈ V (G), G′ v is the subgraph

of G induced by the nodes in V (G′)\Γ[v]. We say that G′v is obtained from G′ by destruction

of v ∈ V (G).

A stable set in G is a subset of mutually nonadjacent nodes in G and a clique is a subset of

pairwise adjacent nodes in G. The cardinality of a stable set of maximum cardinality in G is

denoted by α(G). The stable set polytope in G, STAB(G), is the convex hull of the characteristic

vectors of stable sets in G. The support of a valid inequality of the stable set polytope of a graph

G is the subgraph induced by the nodes with nonzero coefficient in the inequality. A full-support

inequality has G as its support.

If G′ ⊆ G, we may consider every point in STAB(G′) as a point in STAB(G), although they do

not belong to the same space (for the missing nodes, we take direct sums with the interval [0, 1],

since originally STAB(G) ⊆ STAB(G′) ⊕ [0, 1]V (G)\V (G′)). Then, given any family of graphs Fand a graph G, we denote by F(G) the relaxation of STAB(G) defined by

(1) F(G) :=⋂

G′⊆G;G′∈FSTAB(G′).

If FRAC denotes the family of complete graphs of size two, following the definition (1),

the polyhedron FRAC(G) is called the edge relaxation. It is known that G is bipartite if and

only if STAB(G) = FRAC(G). Similarly, if CLIQUE denotes the family of complete graphs,

CLIQUE(G) is the clique relaxation already mentioned and a graph is perfect if and only if

STAB(G) = CLIQUE(G) [14]. Moreover, if OC denotes the family of odd cycles, as a conse-

quence of results in [28] we have the following

Remark 5. If G− v is bipartite for some v ∈ V (G) then STAB(G) = FRAC(G) ∩OC(G).

In [33] Shepherd defined a graph G to be near-bipartite if Gv is bipartite for every v ∈ V (G).

We denote by NB the family of near-bipartite graphs. Since complete graphs and odd cycles are


near-bipartite graphs, it is clear that

NB(G) ⊆ CLIQUE(G) ∩OC(G).

2.2. Minimally imperfect, near-perfect and fs-perfect graphs. Minimally imperfect graphs

are those graphs that are not perfect, but after deleting any node they become perfect. The

Strong Perfect Graph Theorem [13] (also see [11]; and see [12] for the related recognition prob-

lem) states that the only minimally imperfect graphs are the odd cycles and their complements.

Given a graph G it is known that the full-rank constraint∑u∈V

xu ≤ α(G)

is always valid for STAB(G). A graph is near-perfect if its stable set polytope is defined only

by non-negativity constraints, clique constraints and the full-rank constraint [32]. Due to the

results of Chvatal [14], near-perfect graphs define a superclass of perfect graphs and due to the

results in [30], minimally imperfect graphs are also near-perfect. Moreover, every node induced

subgraph of a near-perfect graph is near-perfect [32]. In addition, Shepherd [32] conjectured

that near-perfect graphs could be characterized in terms of certain combinatorial parameters

and established that the validity of his conjecture follows from the Strong Perfect Graph (then

Conjecture, now) Theorem. Therefore,

Theorem 6. ([13],[32]) A graph G is near-perfect if and only if, for every G′ ⊆ G minimally

imperfect, the following two properties hold:

(1) α(G′) = α(G);

(2) for all v ∈ V (G), α(G′ v) = α(G)− 1.

As a generalization of near-perfect graphs, we consider the family of fs-perfect (full-support

perfect) graphs. A graph is fs-perfect if its stable set polytope is defined only by non-negativity

constraints, clique constraints and at most one full-support inequality. Then, every node induced

subgraph of an fs-perfect graph is fs-perfect. We say that a graph is properly fs-perfect if it is

an imperfect fs-perfect graph. Clearly, near-perfect graphs are fs-perfect, but we will see that

fs-perfect graphs define a strict superclass of near-perfect graphs.

2.3. The LS+ operator. In this section, we present a definition of the LS+-operator [28] and

some of its well-known properties when it is applied to relaxations of the stable set polytope of

a graph. In order to do so, we need some more notation and definitions.

We denote by e0, e1, . . . , en the vectors of the canonical basis of Rn+1 where the first coordinate

is indexed zero. Given a convex set K in [0, 1]n,

cone(K) :=

{(x0

x

)∈ Rn+1 : x = x0y; y ∈ K;x0 ≥ 0

}.

Let Sn be the space of n× n symmetric matrices with real entries. If Y ∈ Sn, diag(Y ) denotes

the vector whose i-th entry is Yii, for every i ∈ {1, . . . , n}. Let


M(K) :={Y ∈ Sn+1 : Y e0 = diag(Y ),

Y ei ∈ cone(K),

Y (e0 − ei) ∈ cone(K),∀i ∈ {1, . . . , n}} .

Projecting this polyhedral lifting back to the space Rn results in

LS(K) :=

{x ∈ [0, 1]n :

(1

x

)= Y e0, for some Y ∈M(K)

}.

Clearly, LS(K) is a relaxation of the convex hull of integer solutions in K, i.e., conv(K∩{0, 1}n).

Let Sn+ be the space of n×n symmetric positive semidefinite (PSD) matrices with real entries.

Then

M+(K) := M(K) ∩ Sn+1+

yields the tighter relaxation

LS+(K) :=

{x ∈ [0, 1]n :

(1

x

)= Y e0, for some Y ∈M+(K)

}.

If we let LS0(K) := K, then the successive applications of the LS operator yield LSk(K) =

LS(LSk−1(K)) for every k ≥ 1. Similarly for the LS+ operator. Lovasz and Schrijver proved

that LSn(K) = LSn+(K) = conv(K ∩ {0, 1}n).

In this paper, we focus on the behaviour of the LS+ operator on the edge relaxation of the

stable set polytope. In order to simplify the notation we write LS+(G) instead of LS+(FRAC(G))

and similarly for the successive iterations of it. It is known [28] that, for every graph G,

STAB(G) ⊆ LS+(G) ⊆ TH(G) ⊆ CLIQUE(G)

and

STAB(G) ⊆ LS+(G) ⊆ NB(G).

2.4. LS+-perfect graphs. Recall that a graph G is LS+-perfect if LS+(G) = STAB(G). A

graph that is not LS+-perfect is called LS+-imperfect.

Using the results in [16] and [26] we know that all imperfect graphs with at most 6 nodes

are LS+-perfect, except for the two properly near-perfect graphs depicted in Figure 1, denoted

by GLT and GEMN , respectively. These graphs prominently figure into our current work as the

building blocks of an interesting family of graphs.

From the results in [28], it can be proved that every subgraph of an LS+-perfect graph is

also LS+-perfect. Moreover, every graph for which STAB(G) = NB(G) is LS+-perfect. In

particular, perfect and near-bipartite graphs are LS+-perfect. Recall that in Conjecture 3 we

wonder whether the only LS+-perfect graphs are those graphs G for which STAB(G) = NB(G).

Obviously, G is LS+-perfect if and only if every facet defining inequality of STAB(G) is valid

for LS+(G). In this context, we have Lemma 1.5 in [28] that can be rewritten in the following

way:

Theorem 7. Let ax ≤ β be a full-support valid inequality for STAB(G). If, for every v ∈ V (G),∑w∈V (G−v) ax ≤ β − av is valid for FRAC(G v) then ax ≤ β is valid for LS+(G).


Figure 1. The graphs GLT and GEMN .

In [6] we proved that the converse of the previous result is not true. However, it is plausible

that the converse holds when the full-support valid inequality is a facet defining inequality of

STAB(G). Actually, the latter assertion would be a consequence of the validity of Conjecture 3.

Thus, we present an equivalent formulation of it in the following.

Conjecture 8. If a graph is LS+-perfect and its stable set polytope has a full-support facet

defining inequality, then the graph is near-bipartite.

2.5. Graph operations. In this section, we present some properties of four graph operations

that will be used throughout this paper. Firstly, let us recall the complete join of graphs.

Given two graphs G1 and G2 such that V (G1) ∩ V (G2) = ∅, we say that a graph G is obtained

after the complete join of G1 and G2, denoted G = G1 ∨ G2, if V (G) = V (G1) ∪ V (G2) and

E(G) = E(G1) ∪ E(G2) ∪ {vw : v ∈ V (G1) and w ∈ V (G2)}. A simple example of a join is the

n-wheel Wn, for n ≥ 2 which is the complete join of the trivial graph with one node and the

n-cycle.

It is known that every facet defining inequality of STAB(G1 ∨ G2) can be obtained by the

cartesian product of facets of STAB(G1) and STAB(G2). Hence, odd wheels are fs-perfect but

not near-perfect graphs. Moreover, it is easy to see

Remark 9. The complete join of two graphs is properly fs-perfect if and only if one of them is

a complete graph and the other one is a properly fs-perfect graph.

Also, it is known that

Remark 10. The complete join of two graphs is LS+-perfect if and only if both of them are

LS+-perfect graphs.

Now, let us recall the graph operation, odd-subdivision of an edge [35]. Given a graph G =

(V,E) and e ∈ E, we say that the graph G′ is obtained from G after an odd-subdivision of the

edge e if it is replaced in G by a path of odd length. Next, we consider the k-stretching of a

node which is a generalization of the type (i) stretching operation defined in [26]. Let v be a

node of G with neighborhood Γ(v) and let A1 and A2 be nonempty subsets of Γ(v) such that

A1 ∪ A2 = Γ(v), and A1 ∩ A2 is a clique of size k. A k-stretching of a node v is obtained as

follows: remove v, introduce three nodes instead, called v1, v2 and u, and add an edge between

vi and every node in {u} ∪Ai for i ∈ {1, 2}. Figure 2 illustrates the case k = 1.


A1 A2

v v u1 v2

A1 A2

Figure 2. A 1-stretching operation on node v.

v

v

v

vv

4

v1

56

2

3

v8

v7

v

v

v

vv

4

v1

56

2

3

Figure 3. The graph G′ is obtained from G after the clique subdivision of edge v1v2.

The type (i) stretching operation presented in [26] corresponds to the case k = 0.

We also consider another graph operation defined in [4]. Given a graph G with nodes

{1, . . . , n} and a clique K = {v1, . . . , vs} in G (not necessarily maximal), the clique subdivi-

sion of the edge v1v2 in K is defined as follows: delete the edge v1v2 from G, add the nodes

vn+1 and vn+2 together with the edges v1vn+1, vn+1vn+2, vn+2v2 and vn+ivj for i ∈ {1, 2}and j ∈ {3, . . . , s}. Figure 3 illustrates the clique subdivision of the edge v1v2 in the clique

K = {v1, v2, v6}. Notice that if the clique is an edge this operation reduces to the odd-subdivision

of it.

3. fs-perfection on graphs in Fk

In order to prove Conjecture 8 on fs-perfect graphs, we first consider a minimal structure that

a graph must have in order to be properly fs-perfect and LS+-imperfect. This leads us to define

Fk for every k ≥ 2 as the family of graphs having node set {0, 1, . . . , 2k+1} and such that G−0

is a minimally imperfect graph with 1 ≤ δG(0) ≤ 2k. Let us consider necessary conditions for a

graph in Fk to be fs-perfect.

Theorem 11. [29] Let G ∈ Fk be an fs-perfect graph. Then, the following conditions hold:

(1) α(G) = α(G− 0);

(2) 1 ≤ α(G 0) ≤ α(G)− 1;

(3) the full-support facet defining inequality of STAB(G) is the inequality

(α(G)− α(G 0))x0 +2k+1∑i=1

xi ≤ α(G).


Proof. Let

(2)

2k+1∑i=0

aixi ≤ β

be the full-support facet defining inequality of STAB(G). We may assume that all coefficients

ai, i ∈ {0, . . . , 2k + 1} are positive integers. Clearly,

STAB(G− 0) =

{x ∈ CLIQUE(G− 0) :

2k+1∑i=1

aixi ≤ β

}.

Since G− 0 is a minimally imperfect graph, the inequality∑2k+1

i=1 aixi ≤ β is a positive multiple

of its rank constraint, i.e., there exists a positive integer p such that ai = p for i ∈ {1, . . . , 2k+1}and β = p α(G− 0). Therefore, (2) has the form

(3) a0x0 + p2k+1∑i=1

xi ≤ p α(G− 0).

Observe that there is at least one root x of (3) such that x0 = 1. Clearly, x is the incidence

vector of a stable set S of G such that S − {0} is a maximum cardinality stable set of G 0.

Then,

a0 = p (α(G− 0)− α(G 0)) .

Since a0 ≥ 1, we have α(G 0) ≤ α(G − 0) − 1. Moreover, since δG(0) ≤ 2k, we have

α(G 0) ≥ 1. Therefore, the inequality (3) becomes

(4) (α(G− 0)− α(G 0))x0 +

2k+1∑i=1

xi ≤ α(G− 0).

To complete the proof, we only need to show that α(G) = α(G − 0). Let x be the incidence

vector of a maximum cardinality stable set in G, then

(5) α(G) = x0 +2k+1∑i=1

xi.

Moreover, since α(G− 0)− α(G 0) ≥ 1 and x satisfies (4) we have

α(G) = x0 +2k+1∑i=1

xi ≤ (α(G− 0)− α(G 0))x0 +2k+1∑i=1

xi ≤ α(G− 0).

We have that α(G) ≤ α(G− 0), implying α(G) = α(G− 0). �

As a first consequence of the previous theorem we have:

Corollary 12. Let G ∈ Fk be such that α(G) = 2. Then, G is fs-perfect if and only if

G− 0 = C2k+1 (the complementary graph of C2k+1) and G is near-perfect.

Proof. Assume that G is fs-perfect. By the previous theorem, α(G − 0) = α(G) = 2 and then

G − 0 = C2k+1. Moreover, 1 ≤ α(G 0) ≤ α(G) − 1 = 1 and a0 = 1. Thus, G is near-perfect.

The converse follows from the definition of fs-perfect graphs. �

For k ≥ 2, let Hk denote the graph in Fk having α(Hk) = 2 and δHk(0) = 2k. Using Theorem

6 it is easy to check that Hk is near-perfect. Using Corollary 12, we have the following result:


Corollary 13. Let G ∈ Fk be such that α(G) = 2. Then, G is fs-perfect if and only if G is a

near-perfect edge subgraph of Hk.

Let us now study the structure of fs-perfect graphs G in Fk with stability number at least

3. By Theorem 11, α(G − 0) = α(G) ≥ 3 and G − 0 = C2k+1 with k ≥ 3. Recall that in the

cycle C2k+1 node i is adjacent to node i+ 1 for i ∈ {1, . . . , 2k} and 2k+ 1 is adjacent to node 1.

Clearly, if δ(0) ≤ 2 and 0v ∈ E(G) then G−v is bipartite and, by Remark 5, G is not fs-perfect.

From now on, 3 ≤ s = δ(0) ≤ 2k and Γ(0) = {v1, . . . , vs} such that 1 ≤ v1 < . . . < vs ≤ 2k+1.

Observe that for every i ∈ {1, . . . , s− 1}, the nodes in {w ∈ V (G− 0) : vi ≤ w ≤ vi+1} together

with node 0 form a chordless cycle Di in G. Also, Ds in G is the chordless cycle induced by the

nodes in {w ∈ V (G− 0) : vs ≤ w ≤ 2k + 1 or 1 ≤ w ≤ v1} and node 0. We refer to these cycles

as central cycles of G. Sometimes, we refer to central cycles of length k as k-central cycles. It

is easy to see, using parity arguments, that every G ∈ Fk has an odd number of odd central

cycles. If G has only one odd central cycle, say D1, then G− v1 is bipartite and, by Remark 5,

G is not fs-perfect.

We summarize the previous ideas in the following result:

Lemma 14. Let G ∈ Fk be a fs-perfect graph with α(G) ≥ 3. Then, k ≥ 3, G− 0 = C2k+1 and

G has at least three odd central cycles.

According to the lemma above, we may focus on the structural properties of graphs in Fk

with at least three odd central cycles. Firstly, we have:

Lemma 15. Let G ∈ Fk with k ≥ 3 be such that G − 0 = C2k+1 and G has at least three odd

central cycles. Then, G can be obtained after some odd subdivisions of edges in a graph G′ ∈ Fp

for some 2 ≤ p < k with δG′(0) = δG(0). Moreover,

(1) if every central cycle of G is odd, G′ has one central cycle of length 5 and 2(p−1) central

cycles of length 3;

(2) if G has an even central cycle, every central cycle of G′ has length 3 or 4.

Proof. Let δG(0) = s with s ≥ 3.

(1) If every central cycle of G is odd then s is odd and G has a central cycle with length at

least 5. Let p = s+12 and G′ ∈ Fp with δG′(0) = s such that 0 is adjacent to all nodes

in C2p+1 except to two nodes, e.g., nodes s and s+ 1. Clearly, G′ has one central cycle

of length five, and s − 1 = 2(p − 1) central cycles of length three. Hence G is obtained

after some odd subdivisions of edges of G′.

(2) If G has r ≥ 1 even central cycles then s + r is odd. Let Di with i ∈ {1, . . . , s} be

the central cycles of G. Let p = s+r−12 and G′ ∈ Fp such that δG′(0) = s and for

i ∈ {1, . . . , s} the central cycle D′i in G′ is:

(a) a 3-cycle if Di is odd,

(b) a 4-cycle If Di is even.

It is straightforward to check that G is obtained from G′ after some odd subdivisions of

edges.

�


In addition, we have

Lemma 16. Let G ∈ Fk with k ≥ 3 be such that G − 0 = C2k+1 and δG(0) ≥ 3. Let t(G) be

the number of 3-central cycles and r(G) the number of 4-central cycles in G.

(1) if G has three consecutive 3-central cycles then G can be obtained after the clique subdi-

vision of an edge in a graph G′ ∈ Fk−1 with t(G′) = t(G)− 2 and δG′(0) = δG(0)− 2;

(2) if r(G) ≥ 2 then G can be obtained after the 1-stretching operation on a node in a graph

G′ ∈ Fk−1 with r(G′) = r(G)− 1 and δG′(0) = δG(0)− 1.

Proof. (1) By the hypothesis, we may consider that {2k − 1, 2k, 2k + 1, 1} ⊆ Γ(0). Let

G′ ∈ Fk−1 be a graph having ΓG′(0) = Γ(0) \ {2k, 2k + 1}. Clearly, G′ has t(G) − 2

3-central cycles and δG′(0) = δG(0)− 2. Moreover, it is easy to see that G is the clique

subdivision of the edge in G′ having endpoints {1, 2k − 1}.(2) Since there is a 4-central cycle, without loss of generality, we may assume that the

nodes in {0, 1, 2k, 2k + 1} induce a 4-central cycle in G. Consider G′ ∈ F k−1 such that

ΓG′(0) = ΓG(0) \ {2k}, then G is a 1-stretching of G′ performed at node 1 and where

the new nodes are 2k and 2k + 1. Clearly, r(G′) = r(G)− 1 and δG′(0) = δG(0)− 1.

�

Utilizing the previous lemmas we obtain the following result.

Theorem 17. Let G ∈ Fk with k ≥ 3 and suppose G has at least three odd central cycles. Then,

G can be obtained from GLT or GEMN after successively applying odd-subdivision of an edge,

1-stretching of a node and clique-subdivision of an edge operations.

Proof. Using Lemma 15, by successively performing the odd subdivision of an edge operation,

we may restrict ourselves to consider the following cases:

(a) G has one 5-central cycle and 2(k − 1) 3-central cycles.

(b) G has at least one even central cycle and every central cycle has length 3 or 4.

Consider r(G), the number of 4-cycles in G. If r(G) = 0 then G is a graph described in case (a).

Since k ≥ 3, we have 2(k − 1) ≥ 4, and by Lemma 16 (1), we can conclude that G is obtained

from GLT after successive clique-subdivisions of edges. For graphs in case (b), we have that

r(G) ≥ 1 and then by Lemma 15 (2) we have that 2k = r(G) + δ(G)− 1. If r(G) = 1 and δ(G)

is even, since k ≥ 3 it follows that G has t(G) = δ(G)− 1 ≥ 5 number of 3-cycles. Using Lemma

16 (1) it is not hard to see that G can be obtained from GEMN by successive clique subdivisions

of edges. If r(G) ≥ 2, Lemma 16 (2) implies that G can be obtained by successive 1-stretching

of nodes from a graph G′ ∈ Fk′ with r(G′) = 1 and 2k′ = δ(G′). If 2k′ = 4, then G′ = GEMN ;

otherwise, we can refer to the previous case and the proof is complete. �

4. The conjecture on fs-perfect graphs

In this section, we prove the validity of Conjecture 8 on the family of fs-perfect graphs. We

start by proving it on graphs in the family Fk with k ≥ 2. Recall that when G ∈ Fk is a

fs-perfect graph with α(G) = 2, Corollary 13 states that G is a near-perfect edge subgraph of

Hk. Observe that H2 = GEMN and then, due to the results in [16], Hk is LS+-imperfect when

k = 2. Next, we prove the imperfection property for the whole family of graphs Hk.


Theorem 18. For k ≥ 2, the graph Hk is LS+-imperfect.

In order to ease the reading of this paper we postpone the proof of Theorem 18 to Section 5.

Regarding the behaviour of the LS+ operator on edge subgraphs, we have the following result:

Lemma 19. Let GE be an edge subgraph of G and ax ≤ β be a valid inequality for STAB(GE).

Then, if ax ≤ β is not valid for LSr+(G), then STAB(GE) 6= LSr+(GE).

Proof. Clearly, by definition LSr+(G) ⊆ LSr+(GE). Thus, if there exists x ∈ LSr+(G) such that

ax > β then ax ≤ β is not valid for LSr+(GE). Moreover, by hypothesis, ax ≤ β is valid for

STAB(GE) and the result follows. �

As a consequence of the above, we have:

Theorem 20. Let G ∈ Fk be an fs-perfect graph with α(G) = 2. Then, G is LS+-imperfect.

Proof. By Corollary 13 we know that G is a near-perfect edge subgraph of Hk. Theorem 18

states that Hk is LS+-imperfect; thus, the full rank constraint is not valid for LS+(Hk). Since

α(Hk) = α(G) = 2, the full rank constraint is not valid for LS+(G) and G is LS+-imperfect. �

Let us consider the fs-perfect graphs G in Fk with α(G) ≥ 3. Due to the structural charac-

terization in Theorem 17, we are interested in the behavior of the LS+-operator under the odd

subdivision of an edge, k-stretching of a node and clique subdivision of an edge operations. In

this context, a related earlier result is:

Theorem 21 ([26]). Let G be a graph and r ≥ 1 such that LSr+(G) 6= STAB(G). Further

assume that G is obtained from G by using the odd subdivision operation on one of its edges.

Then, LSr+(G) 6= STAB(G).

Concerning the remaining operations, we present the following results whose proofs are in-

cluded in Section 5 for the sake of clarity.

Theorem 22. Let G be a graph and r ≥ 1 such that LSr+(G) 6= STAB(G). Further assume that

G is obtained from G by using the k-stretching operation on one of its nodes. Then, LSr+(G) 6=STAB(G).

Theorem 23. Let G be a graph and r ≥ 1 such that LSr+(G) 6= STAB(G). Further assume

that G is obtained from G by using the clique subdivision operation on one of its edges. Then,

LSr+(G) 6= STAB(G).

In summary, we can conclude that the odd-subdivision of an edge, the 1-stretching of a node

and the clique-subdivision of an edge are operations that preserve LS+-imperfection. Then, the

behavior of these operations under the LS+ operator together with the fact that graphs GLT

and GEMN are LS+-imperfect, Lemma 14 and Theorem 17 allow us to deduce:

Theorem 24. Let G ∈ Fk be an fs-perfect graph with α(G) ≥ 3. Then, G is LS+-imperfect.

Finally, we are able to present the main result of this contribution.

Theorem 25. Let G be a properly fs-perfect graph which is also LS+-perfect. Then, G is the

complete join of a complete graph (possibly empty) and a minimally imperfect graph.


Proof. Since G is a properly fs-perfect graph, G has a (2k+1)-minimally imperfect node induced

subgraph G′. If G′ = G the theorem follows. Otherwise, let v ∈ V (G) \V (G′) and let Gv be the

subgraph of G induced by {v}∪V (G′). Clearly, Gv is properly fs-perfect as well as LS+-perfect.

Then, by Theorem 20 and Theorem 24, Gv /∈ Fk. So, δGv(v) = 2k + 1 and Gv = {v} ∨ G′.Therefore, if G′′ is the subgraph of G induced by V (G) − V (G′), G = G′ ∨ G′′. By Remark 9,

G′′ is a complete graph and by Remark 10 the result follows. �

Since complete joins of complete graphs and minimally imperfect graphs are near-bipartite,

they satisfy NB(G) = STAB(G). Therefore, based on the results obtained so far, we can conclude

that Conjecture 8 holds for fs-perfect graphs.

5. Results concerning the LS+-operator

In this section we include the proofs of some results on the LS+-operator that were stated

without proof in the previous sections.

5.1. The LS+-imperfection of the graph Hk. Recall that V (Hk) = {0, 1, . . . , 2k+ 1}, Hk−0 = C2k+1 and δ(0) = 2k. Without loss of generality, we may assume that the node 2k +

1 in Hk is the only one not connected with node 0. Let us introduce the point x(k, γ) :=1

2k+2+γ (2, 2, . . . , 2, 4)> ∈ R2k+2 where for i ∈ {1, . . . , 2k + 2}, the i-th component of x(k, γ)

corresponds to the node i − 1 in Hk. In what follows, we show that x(k, γ) ∈ LS+(Hk) \STAB(Hk) for some γ ∈ (0, 1) thus proving Theorem 18. We first consider βk = 1

2k+2 , γ ∈ (0, 1)

and the (2k + 3)× (2k + 3) matrix given by

Y (k, γ) :=

(2k + 2 + γ) 2 2 2 2 2 · · · 2 4

2 2 0 0 0 0 · · · 0 2

2 0 2 1− βk 0 0 · · · 0 1 + βk

2 0 1− βk 2 1 + βk 0 · · · 0 0

2 0 0 1 + βk 2 1− βk · · · 0 0

2 0 0 0 1− βk 2 · · · 0 0

......

......

......

. . ....

...

2 0 0 0 0 0 · · · 2 1 + βk

4 2 1 + βk 0 0 0 · · · 1 + βk 4

.

Lemma 26. For k ≥ 3, there exists γ ∈ (0, 1) such that Y (k, γ) is PSD.

Proof. Let us denote by Y (k) the (2k+2)×(2k+2) submatrix of Y (k, γ) obtained after deleting

the first row and column. Also consider Y (k), the Schur Complement of the (1, 1) entry of Y (k),


then

Y (k) =

2 1− βk 0 0 · · · 0 1 + βk

1− βk 2 1 + βk 0 · · · 0 0

0 1 + βk 2 1− βk · · · 0 0

0 0 1− βk 2 · · · 0 0

......

......

. . ....

...

0 0 0 0 · · · 2 1 + βk

1 + βk 0 0 0 · · · 1 + βk 2

.

Claim 27. For every k ≥ 2, Y (k) is positive definite.

Proof. Let us first show that Y (k) is positive definite. For this purpose, we only need to verify

that every leading principal minor of Y (k) is positive. Let us define A0(k) := 1, B0(k) := 2 and

for ` ≥ 1,

A`(k) := det

2 1− βk 0 0 · · · 0 0

1− βk 2 1 + βk 0 · · · 0 0

0 1 + βk 2 1− βk · · · 0 0

0 0 1− βk 2 · · · 0 0

......

......

. . ....

...

0 0 0 0 · · · 2 1− βk

0 0 0 0 · · · 1− βk 2

,

where the matrix in the definition is 2`× 2` and

B`(k) := det

2 1− βk 0 0 · · · 0 0

1− βk 2 1 + βk 0 · · · 0 0

0 1 + βk 2 1− βk · · · 0 0

0 0 1− βk 2 · · · 0 0

......

......

. . ....

...

0 0 0 0 · · · 2 1 + βk

0 0 0 0 · · · 1 + βk 2

,

where the matrix in the definition is (2` + 1) × (2` + 1). Using the determinant expansion on

A`(k) and B`(k) we have that for every ` ≥ 1,

A`(k) = 2B`−1(k)− (1− βk)2A`−1(k),

B`(k) = 2A`(k)− (1 + βk)2B`−1(k),

and

det(Y (k)

)= 2

[Ak(k)− (1 + βk)

2Bk−1(k) + (1− βk)k(1 + βk)k+1]

= 2[Bk(k)−Ak(k) + (1− βk)k(1 + βk)

k+1].(6)


Using these recursive formulas, we find that for every ` ≥ 1,

A`(k) = (2`+ 1− βk) (1− βk)`−1(1 + βk)`,(7)

B`(k) = (2`+ 2) (1− βk)`(1 + βk)`.(8)

Note that A`(k) and B`(k) are positive for every ` and every k. Moreover, every leading principal

minor of Y (k), except itself, is either A`(k) orB`(k) for some `. Using (6), (7) and (8) we compute

det(Y (k)

)= 2

[2− (2k + 1)βk − β2

k

](1− βk)k−1(1 + βk)

k(9)

which is indeed positive. Since we verified that every principle minor of Y (k) is positive, we

deduce that Y (k) is positive definite. Finally, after applying the Schur Complement Lemma we

have that Y (k) is positive definite. �

Using this claim we have:

Claim 28. Let u be the (unique) vector such that

(10) Y (k)u = 2(1 + e2k+2).

Then Y (k, γ) is PSD if and only if γ ≥ 1− βku2k+2.

Proof. Using the Schur Complement Lemma for Y (k) we have that Y (k, γ) is PSD if and only

if

Y (k)− 4

2k + 2 + γ(1 + e2k+2)(1 + e2k+2)> is PSD.

Using the automorphism[Y (k)

]−1/2·[Y (k)

]−1/2of the PSD cone, the latter is true if and only

if the following matrix

(11) I − 4

2k + 2 + γ

[Y (k)

]−1/2(1 + e2k+2)(1 + e2k+2)>

[Y (k)

]−1/2

is PSD. Since4

2k + 2 + γ

[Y (k)

]−1/2(1 + e2k+2)(1 + e2k+2)>

[Y (k)

]−1/2

is a rank one matrix, using (10) we have that the matrix in (11) is PSD if and only if

(12) 1 ≥ 4

2k + 2 + γ(1 + e2k+2)>[Y (k)]−1(1 + e2k+2) =

2(1 + e2k+2)>u

2k + 2 + γ.

To conclude the above, we used the observation that for every h ∈ Rn \ {0},

I − hh> is PSD iffh>

‖h‖2

(I − hh>

) h

‖h‖2≥ 0.

To see the latter, one can apply a spectral decomposition to the matrix(I − hh>

), where one

of the eigenvectors is h‖h‖2 .

Now, using the definition of Y (k) we have that

Y (k)1 = 4(1) + (4 + 2βk)e2k+2 = 2Y (k)u+ 2βke2k+2,

and then

(13) u =1

21− βk[Y (k)]−1e2k+2.


Therefore,

2(1 + e2k+2)>u = 2(1 + e2k+2)>(

1

21− βk[Y (k)]−1e2k+2

)= (2k + 3)− 2βk(1 + e2k+2)>[Y (k)]−1e2k+2

and again using (10), we obtain

(14) 2(1 + e2k+2)>u = (2k + 3)− βku2k+2.

Hence, using (12) and (14) we can conclude that the matrix in (11) is PSD if and only if

1 ≥ (2k + 3)− βku2k+2

2k + 2 + γ

or equivalently, if and only if

γ ≥ 1− βku2k+2.

�

By the previous claims, to prove that Y (k, γ) is PSD for some γ ∈ (0, 1), it suffices to prove

that there exists γ ∈ (0, 1) such that

γ ≥ 1− βku2k+2,

where u is the unique solution of (10). Thus, as long as u2k+2 > 0, we may have γ < 1 as

desired. Using (13), we have that

u2k+2 = e>2k+2u =1

2− βke>2k+2Y (k)−1e2k+2.

Based on the above expression, we may interpret e>2k+2Y (k)−1e2k+2 as the (2k + 2)nd entry of

the unique solution h of the linear system Y (k)h = e2k+2. Now using Cramer’s rule to express

this entry as a ratio of two determinants and the definitions of Y , Y , and Ak, we conclude

u2k+2 =1

2− βk

Ak(k)

det(Y (k)

) .Then, using the identities (7) and (9), we obtain

u2k+2 =1

2

[1− βk

2k + 1− βk2− (2k + 1)βk − β2

k

].

The numerator of the last ratio above is strictly less than (2k+1) and the denominator is strictly

larger than one. Whence, u2k+2 >1

4(k+1) > 0. This completes the proof. �

Utilizing the previous lemma we are able to prove Theorem 18.

Proof of Theorem 18. Recall, we may assume that in Hk the node 2k + 1 is not connected to

node 0. Let γ ∈ (0, 1) and x(k, γ) = 12k+2+γ (2, 2, . . . , 2, 4)> ∈ R2k+2 where the i-th component

of x(k, γ) corresponds to node i−1 in Hk for i ∈ {1, . . . , 2k+2}. Let Y ∗(k, γ) := 12k+2+γY (k, γ).

Then, Y ∗(k, γ) is a symmetric matrix that clearly satisfies that Y ∗(k, γ)e0 = diag(Y ∗(k, γ)) ∈FRAC(Hk). Moreover, it is not hard to check that, for i ∈ {1, . . . , 2k + 2},

Y ∗(k, γ)ei ∈ cone(FRAC(Hk)) and Y ∗(k, γ)(e0 − ei) ∈ cone(FRAC(Hk)).


This proves that Y ∗(k, γ) ∈ M(Hk). By the previous lemma, there exists γ ∈ (0, 1) for which

Y ∗(k, γ) ∈M+(Hk). Hence, x(k, γ) ∈ LS+(Hk). It only remains to observe that x(k, γ) violates

the rank inequality of Hk. Thus, x(k, γ) /∈ STAB(Hk). �

5.2. Operations that preserve LS+-imperfection. Firstly, we prove Theorem 22 on the

k-stretching operation for k ≥ 1, already stated in Section 4. Actually, we will see that the same

proof given in [26] for the case k = 0 can be used for the case k ≥ 1. Assume that G is obtained

from G after the k-stretching operation on node v and let u, v1 and v2 be as in the definition of

the operation in Section 2.5. For any x ∈ RV (G), we write x =

(x

xv

)where x ∈ RV (G−v).

For the case k = 0, the authors in [26] prove that if a point x =

(x

xv

)∈ LSr+(G) then the

point x given by

xw =

{xw if w ∈ {u, v1, v2},xw otherwise,

satisfies x ∈ LSr+(G). In order to do so they prove that if

Y =

1 x> xv

x X y

xv y> xv

∈M+(LSr−1+ (G))

then

Y =

1 x> xv xv (1− xv)

x X y y x− y

xv y> xv xv 0

xv y> xv xv 0

(1− xv) (x− y)> 0 0 (1− xv)

∈M+(LSr−1

+ (G)).

On the other hand, they show that if∑

j∈V (G) ajxj ≤ β is a valid inequality for STAB(G),

defining β = β + av and

aj =

av if j ∈ {v1, v2, u},

aj otherwise,

the inequality∑

j∈V (G) ajxj ≤ β is valid for STAB(G). Moreover, if x∗ violates∑

j∈V (G) ajxj ≤β then x∗ violates

∑j∈V (G) ajxj ≤ β.

Proof of Theorem 22. It is enough to observe that Y ∈M+

(LSr−1(G)

)and the inequality∑

j∈V (G) ajxj ≤ β is valid for STAB(G) even for the case that G is obtained after the k-stretching

on node v in G, for k ≥ 1. �


Let us now consider the clique-subdivision operation defined in [4]. For x ∈ Rn, let x ∈ Rn+2

such that xi = xi for every i ∈ {1, . . . , n}, xn+1 = x2 and xn+2 = x1, and write x =

x

x2

x1

.

In [4] the authors prove that if G is obtained from G by the clique subdivision of the edge

v1v2 in the clique K and x ∈ LSk+(G) then x ∈ LSk(G). In order to do so, they show that if

Y e0 =

(1

x

)for

(15) Y =

1 x1 x2 x>

x1 x1 0 y>1x2 0 x2 y>2

x y1 y2 X

∈M(LSr−1(G))

then

(16) Y =

1 x1 x2 x> x2 x1

x1 x1 0 y>1 0 x1

x2 0 x2 y>2 x2 0

x y1 y2 X y2 y1

x2 0 x2 y>2 x2 0

x1 x1 0 y>1 0 x1

∈M(LSr−1(G)).

Proof of Theorem 23. It is enough to observe that if the matrix Y is PSD then so is the matrix

in (16). �

6. Conclusions and further results

In this work, we face the problem of characterizing the stable set polytope of LS+-perfect

graphs, a graph class where the Maximum Weight Stable Set Problem is polynomial-time solv-

able. This class strictly includes many well-known graph classes such as perfect graphs, t-perfect

graphs, wheels, anti-holes, near-bipartite graphs and the graphs obtained from various suitable

compositions of these. The stable set polytope of either a perfect or a near-bipartite graph only

needs the inequalities associated with the stable set polytopes of its near-bipartite subgraphs.

We have conjectured that the same holds for all LS+-perfect graphs. In this paper, we prove

the validity of this conjecture for fs-perfect graphs, a superclass of near-perfect graphs. More-

over, if FS denotes the class of fs-perfect graphs, using the definition in (1), we actually prove

that the conjecture holds for a superclass of fs-perfect graph defined as those graphs for which

FS(G) = STAB(G). Observe that the graph in Figure 4 satisfies FS(G) = STAB(G) and it is

not fs-perfect.

Also, the results used in the proof of the Theorem 25 allow us to conclude the following:


Figure 4. A graph G satisfying FS(G) = STAB(G) which is not fs-perfect.

Corollary 29. Let G be a graph such that V (G) = {0, 1, . . . , 2k + 1} with k ≥ 2 and G − 0 is

minimally imperfect. Then:

• If G − 0 = C2k+1 then G is LS+-perfect if and only if either δG(0) ≥ 2 and G has only

one odd central cycle or δG(0) ∈ {0, 1, 2k + 1}.• If G− 0 = C2k+1 with α(G) = 2 then G is LS+-perfect if and only δG(0) = 2k + 1.

From the above characterization, we identify some of the forbidden structures in the family

of LS+-perfect graphs:

Corollary 30. Let G be an LS+-perfect graph. Then, there is no subgraph G′ of G such that

• G′ − v0 = C2k+1, 2 ≤ δG′(v0) ≤ 2k and G′ has at least two odd central cycles, or

• G′ − v0 = C2k+1 and k + 1 ≤ δ′G(v0) ≤ 2k and α(G′) = 2,

for some v0 ∈ V (G′) and k ≥ 2.

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